Amount of Propane Required to Melt 1L Aluminum

[Voltux]

I started off yesterday learning about specific heats, and measuring my ice cubes to figure out the correct number to cool my tea to a proper drinking temperature.

Now today I was curious how much propane would be required to melt 1L of Aluminum since I'm also interested in casting Aluminum. I was hoping that someone could check my work, and then tell me how I can get more "Real world" results, or if this would even be practical at all for my level of math/physics/chem understanding.

Also, if you have any concepts you think I should learn next I would appreciate any suggestions. I have some Physics, and Chemistry books that I am reading for furthering my education.

Aluminum
Latent Heat of Fusion: 398 J/g
Specific Heat: 0.91 J/g°C
Density: 2.70 g/cm^3
Melting Point: 660.32°C
Ambient Temperature: 20°C

Mass of 1 Liter Aluminum (Mass = Density * Volume) (1mL = 1cm^3)
2.70g/cm^3 * 1000mL = 2700g

Required Energy for +640.32°C (Mass * Specific Heat * Temperature Difference)
2700 * 0.91 * (660.32-20)
2700 * 0.91 * 640.32
2457 * 640.32 = 1,573,266.24

1,573,266.24 J Required to Heat 20°C Aluminum to 660.32°C

Required Energy to Change Solid -> Liquid (Mass * Latent Heat of Fusion)
2700 * 398 = 1,074,600 J

Total Energy to Heat, and Melt Aluminum (Raise Temp + State Change)
1,573,266.23 + 1,074,600 = 2,647,866.23 J

2,647,866.23 J
735.5184 Wh (Joules/3600)
2509.6929443 BTU (Joules/1055.05585262)

Propane
Molar Mass: 44.1g/mol
Energy Density: 2220kJ/mol
Liquid Density at 25°C: 0.493g/cm^3
Mass Per Liter: 493g/Liter

Moles Per Liter (Mass Per Liter/Molar Mass)
493/44.1 = 11.17913832mols

Energy Density per Liter (Mol * Energy Density)
11.17913832 * 2,220,000 = 24817687.0704J

Amount of Aluminum Melted per Liter of Propane (Energy per Liter Propane/Energy to Melt Aluminum)
24817687.0704/2,647,866.23 = 9.3727118044L Melted per Liter of Propane

 

[Vanadium 50]

The first thing you could do to get more "real world" results is to use significant figures. 24817687.0704? Good heavens!

 

[mfb]

Well, in real setups, you don't get 100% efficiency. You will heat a lot of air and other parts of your experiment as well. The efficiency really depends on the setup.

Working with units everywhere and with less [strike]in[/strike]significant figures would be useful.

 

[Catria]

9.372L of aluminum melted using just 1L of propane? Holy moly... Doing it for real will likely result in maybe 20-25% efficiency; you'd be lucky to be able to melt 3L with that much propane for real.